• 工时统计的sql练习--包含时间处理


    //按月统计,除去周末的考勤,(工时,请假,缺勤)

    --建表sql 创建[dbo].[AbsenceHourld]
    CREATE TABLE [dbo].[AbsenceHourld](
    [id] [int] NULL,
    [date] [datetime] NULL,
    [name] [varchar](50) NULL,
    [hours] [int] NULL
    ) ON [PRIMARY]
    --创建[dbo].[ManHour]
    CREATE TABLE [dbo].[ManHour](
    [id] [int] NULL,
    [date] [datetime] NULL,
    [name] [varchar](50) NULL,
    [hours] [int] NULL
    ) ON [PRIMARY]
    --建表sql
    --统计所有员工-- 按天统计考勤数据
    select a.name as 姓名,
    CONVERT(varchar(100), a.[date] ,111) as 时间天,
    sum(a.[hours])as 工作时间,
    sum(isnull(b.[hours],0)) as 请假时间 ,
    8-sum(a.[hours])+sum(isnull(b.[hours],0)) as 缺勤时间
    from ManHour a
    left join AbsenceHourld b
    on a.name=b.name
    where datepart(weekday, a.[date]) not in (6,7)
    group by CONVERT(varchar(100), a.[date] ,111) ,a.name

    -- 创建返回某月去掉周末的有效工时天数——函数
    alter FUNCTION 返回减周末后月的天数(@date varchar(8))
    RETURNS INT
    AS
    BEGIN
    declare @fistM varchar(8) --月的第一天
    declare @YY char(4),@MM char(2) --年,月
    set @YY=LEFT(@date,4)
    set @MM=dateadd(mm,1,@date) --这种截取会让月份是0开头
    set @fistM=@YY+@MM+'01'
    DECLARE @edate varchar(8),@monthCount INT ,@weekCount int --月的结束时间,月的天数,周末的天数
    --取出月的天数 @monthCount
    SET @edate =CONVERT(VARCHAR(8), dateadd(D,-1,@fistM),112)
    SET @monthCount =CAST(datepart(day,@edate) AS INT)
    --月的周末数 @weekCount
    set @weekCount=0
    declare @wdate varchar(8), @temp_time varchar(8),@WM varchar(6) --周计算(月)第一天,中间变量
    set @WM=LEFT(@date,6)
    set @wdate=@WM+'01'
    set @temp_time=@wdate
    while left(@wdate,6)=left(@temp_time,6)
    begin
    IF DATEPART(WEEKDAY,@temp_time)=1 OR DATEPART(WEEKDAY,@temp_time)=7 --第一天是星期几
    set @weekCount=@weekCount+1
    set @temp_time=convert(varchar(8),dateadd(D,1,@temp_time),112)
    end
    return @monthCount-@weekCount
    END

    select dbo.返回减周末后月的天数('20150105')

    --统计所有员工 --月度考勤数据
    select a.name as 姓名,
    convert(char(4),year(a.[date]))+'年'+convert(char(2),month(a.[date]))+'月' as 时间,
    sum(a.[hours])as 工作时间,
    sum(isnull(b.[hours],0)) as 请假时间,
    8*dbo.返回减周末后月的天数(CONVERT(varchar(12),min(a.[date]), 112 )) --调用函数取得
    -sum(a.[hours])+sum(isnull(b.[hours],0)) as 缺勤时间
    from ManHour a
    left join AbsenceHourld b
    on a.name=b.name
    where datepart(weekday, a.[date]) not in (6,7)
    group by month(a.[date]),a.name,year(a.[date])

  • 相关阅读:
    C#中ToString格式大全
    mysql事务
    Mac eclipse 启动卡住
    Mac 安装zkdash
    Mac 安装SecureCRT
    java多线程、并发系列之 (synchronized)同步与加锁机制
    jvm 年轻代
    查看日志技巧
    which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by
    tomcat限制内存
  • 原文地址:https://www.cnblogs.com/zoumin123/p/4903235.html
Copyright © 2020-2023  润新知